@Alizter: The CRT gives all the solutions - they're exactly $a + n 13 \cdot 14 \cdot 15$ for each $n \in \mathbb Z$. Pretty sure there are no answers with magnitude less than 1000.
Here is the question What is the smallest positive integer that can be expressed as the sum of eleven consecutive positive integers, the sum of twelve consecutive positive integers, and the sum of thirteen consecutive positive integers?
So i assume using triangle numbers $$T_{n_1+12}-T_{n_1}=T_{n_2+13}-T_{n_2}=T_{n_3+14}-T_{n_3}=a$$
@alizter: No problem. According to mathematica our answer was correct for the modified problem a == Sum[k, {k, m, m + 12}] == Sum[k, {k, n, n + 13}] == Sum[k, {k, p, p + 14}]
@robjohn I'm glad my bounty goes to you. :-) (I still wonder why I missed that point in achille's answer ... I think sometimes our mind makes us believe things are OK when one strongly believes in the power of a person of making right things)
It's something there that makes you unable to see the mistakes.
@robjohn or maybe because of too less sleep (5 hours/ night)
@Alizter: Given $z = (12-1)(12+1)$ you're looking for $y$ such that $zy \equiv 1$, i.e. $y \equiv z^{-1}$ - that is, $y$ is the multiplicative inverse of $z$ when working modulo 12.
@Chris'ssis Whee! thanks! I wondered how he got a better estimate on the lower bound, and when I looked I realized it was too good. It essentially implies that $a_n=n/2$, which is obviously false.
@Chris'ssis Ouch! I didn't notice, until I just looked, that achille deleted his answer. He could have found a way to get the lower bound. Maybe there is no other easy way; I haven't looked.
Guys two questions, is an element of surface defined for a Mobius strip, surely if we can describe it parametrically we can at least have a go, and just choose limits very carefully.
Also is this a joke: let $f(x)=c$ and $g(x)=e^x$ the product function, $fg$ is afraid of which differential operator.
@Alizter Sorry, I was out getting plumbing supplies. Now that i look at the equations I posted, if multiplied by $xyz$ they become $x-2x^2=y-2y^2=z-2z^2$...
@skullpatrol That's right. The big problem is that no one teaches you how to think on your own without help ... and many kids believe that their teachers are endowed with some magical powers and that's why they are able to solve all kinds of problems. No one tells them that the teachers worked many years to get a certain level in math, that there is about a huge amount of work.
@PeterTamaroff some time ago I entered a university with a perfect score, only problems from high school (but hard ones), and I can say that during the learning period I fell in love with those questions, very nice questions. (I dropped that university after a while)
@what'sup That is some good work. I am kinda busy at the moment getting ready for first day of school tomorrow. Tomorrow when I get back I will have a proper look at it. Thank you for the help!
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@KarlKronenfeld because we are embarresed to tell little kids we do not understand such a simple concept and hope to find a solution before we have kids ourselves :D
@Chris'ssis If you want to learn from the best, get the book "Problems and Theorems in Analysis" and "Inequalities". The first one is from Polya and Szego, the second is Polya, Littlewood and Hardy. If you cannot find them, let me know.
@PeterTamaroff yes, it works with the help of Mathematica. The differentiation under the integral sign seems to work fine. I'll also think to change its appearance and use some results met in complex analysis.
I used the case $$\sum_{n=1}^\infty \frac{n}{e^{2\pi n}-1}=\frac{1}{24}-\frac{1}{8\pi}$$ Along with some q series identities to get, $$\frac{1}{\pi}=\frac{1}{3}-8\sum_{n=1}^\infty e^{-2\pi n^2}n\coth(\pi n)-2\sum_{n=1}^\infty e^{-2\pi n^2}\text{csch}(\pi n)^2$$
@PeterTamaroff I'm sure the result can be got by some elegant manipulations that also involve the use of the integration by parts. Now I need to get some sleep but I'll compute it in more ways.
Let $p^*_n$ be the $n$ th element of a subset of primes such that $p^*_{n+1}>p^*_n$ and $p^*_n < O((n+2) ln((n+2))^3)$. Define $f(z)$ as the analytic continuation of $\prod_{n>0} (1+\dfrac{1}{p^{*^z}_n-1})$.
The analytic continuation is to the largest possible domain.
If there is no natural boun...
Quick Q, what rule in particular causes $7^{logn^2}$ to dominate $logn^{logn}$? Exponents with base exponents dominate standard exponents? Or which rule applies? Thanks!
